In general exponentiation is not commutative. However most people are aware of the pair 2 and 4 as a pair of unequal numbers for which exponentiation is commutative.
Show that there are an infinite number of pairs of unequal numbers for which exponentiation is commutative.
Go further and show that there are an infinite number of pairs of unequal rational numbers for which exponentiation is commutative.
What is sought is real numbers where a^b = b^a or equivalently
a^(1/a) = b^(1/b).
I recall asking myself when x^(1/x) was a maximum back when I first got a graphing calculator. The graph of y=x^(1/x) for x>0 starts with an open dot at (0,0) rises quickly through (1,1) then levels off and reaches a max ar (e, e^(1/e)). From there it decreases towards a horizontal asymptote y=1.
Any horizontal line between y=1 and y=e^(1/e)=1.4446679... will hit the graph twice. Once with x between 1 and e and the other with x greater than e. These are the values of a and b and there are infinitely many of them.
Example a=1.5
The horizontal line y=1.5^(1/1.5) =1.3103707... hits the graph twice, once at x=1.5 and again at 7.4087647...
so b=7.4087647...
a^b=b^a=20.16595073...
These numbers are both real, but b is not rational (as far as I can tell).
I haven't found any rational solutions besides the integer pair given. If I had to guess, I'd think there aren't any others, but the problem asks to show there are.
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Posted by Jer
on 2017-07-07 12:20:19 |
Real solutions
